Question

An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.7 m/s. A child on a bridge, 72 m above the river, sees the raft in the river below and attempts to drop a small stone onto the raft. The child releases the stone from rest. In order for the stone to hit the raft, what must be the horizontal distance between the raft and the bridge when the child releases the stone?

Answer 1

first we need to calculate time taken by the stone to drop in boat, and then using this time we will calculate required distance between raft and the bridge.

Now Using 2nd kinematic equation:

h = U*t + (1/2)*a*t^2

h = height of bridge = 72 m (downward positive)

U = initial velocity of stone = 0 m/sec

a = acceleration = g = 9.81 m/sec^2

So Using these values:

72 = 0*t + (1/2)*9.81*t^2

t = sqrt (2*72/9.81)

t = 3.83 sec

Now using this time, distance between bridge and raft should be

distance = speed of raft*time available

speed of raft = 5.7 m/sec

So,

d = distance between raft and the bridge = (5.7 m/sec)*(3.83 sec)

d = 21.83 m

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